What amount of time it will take to pass $10^{19}$ electrons through a point inside the circuit, if a constant current of 8 A is flowing in it?

8 \times 10^{- 19}

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5 s

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1.6 \times 10^{19}

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0.2 s

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Solution

The correct option is D 0.2 s
Given:
Number of electrons $n = 10^{19}$ , charge of each electron $= 1.6 \times 10^{- 19}$
Current $I = 8 A .$
We know that, q=ne
Also, I = $\frac{q}{t}$
Hence t = $\frac{n e}{I}$ = $\frac{10^{19} \times 1.6 \times 10^{- 19}}{8}$ = 0.2 s

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For what time period should a flow of $10^{19}$ electrons be maintained between two points A and B to have a constant current of $8 A$ between them?

How long should a flow of $10^{19}$ electrons be maintained between two points A and B to have a constant current of 8 A flowing between them?

The charge of an electron is known to be $1.6 \times 10^{- 19} C$ . In a circuit, the current flowing is $1 A$ . How many electrons will be flowing through the circuit per second?

How long should a flow of $10^{19}$ electrons be maintained between two points A and B to have a constant current of 8 A flowing between them?

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What amount of time it will take to pass 1019 electrons through a point inside the circuit, if a constant current of 8 A is flowing in it?

The correct option is D 0.2 sGiven: Number of electrons n=1019, charge of each electron =1.6×10−19 Current I=8 A. We know that, q=ne Also, I = qt Hence t = neI = 1019×1.6×10−198 = 0.2 s

What amount of time it will take to pass $10^{19}$ electrons through a point inside the circuit, if a constant current of 8 A is flowing in it?

The correct option is D 0.2 s
Given:
Number of electrons $n = 10^{19}$ , charge of each electron $= 1.6 \times 10^{- 19}$
Current $I = 8 A .$
We know that, q=ne
Also, I = $\frac{q}{t}$
Hence t = $\frac{n e}{I}$ = $\frac{10^{19} \times 1.6 \times 10^{- 19}}{8}$ = 0.2 s